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In describing the propagation of light as a wave we need to understand: wavefronts: a surface passing through points of a wave that have the same phase.

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Presentation on theme: "In describing the propagation of light as a wave we need to understand: wavefronts: a surface passing through points of a wave that have the same phase."— Presentation transcript:

1 In describing the propagation of light as a wave we need to understand: wavefronts: a surface passing through points of a wave that have the same phase. rays: a ray describes the direction of wave propagation. A ray is a vector perpendicular to the wavefront.

2 Reflection and Refraction When a light ray travels from one medium to another, part of the incident light is reflected and part of the light is transmitted at the boundary between the two media. The transmitted part is said to be refracted in the second medium. reflected rayincident ray

3 The Law of Reflection For specular reflection the incident angle  i equals the reflected angle  r   i    r The angles are measured relative to the normal, shown here as a dotted line.

4 Types of Reflection specular reflection diffuse reflection

5 Mirrors reflect light and allow us to see ourselves. O is the object or its coordinate i is the image or its coordinate p is the distance of the object to a mirror, refracting surface or lens q (or i) is the distance of the image to a mirror, refracting surface or lens h is the object height h’ is the image height lateral magnification is the ratio of image height to object height an image is real if the light converges to form the image in space an image is virtual if the light appears to come from a place where it cannot

6 Plan Mirrors Illustrating formation of an image by a plane mirror. Since QR is common to both triangle PQR and triangle P’QR and  is is the same angle at vertex P and vertex P’ the right triangles are congruent, and p = - q, also h = h’ or the lateral magnification (M) is +1. The image is upright, the same size and left-right reversed

7 Images from Mirrors

8 Spherical Mirrors Definitions for the following terms –Center of curvature (C) –Radius of curvature (R or r) –Principle Axis (or symmetry axis) –Vertex (V)

9 Spherical Mirrors When parallel rays (e.g. rays from a distance source) are incident upon a spherical mirror, the reflected rays intersect at the focal point f, a distance R/2 from the mirror.

10 The image formeation

11 Sign Convention

12 Spherical Mirrors Convex mirrors: virtual images only

13 Spherical Mirrors Focus and focal length

14 Spherical Mirrors

15 Light inside a medium Law of Refraction (Snell’s Law) n 1 sin  1 = n 2 sin  2 sin  c = n 2 / n 1 Critical Angle Required : n 1 > n 2  1 >  c

16 Spherical Refracting Surfaces Flat refracting surfaces and apparent depth

17 Thin Lenses Two spherical refracting surfaces back to back Thickness of lens is small (negligible)

18 Thin Lenses

19

20 Converging and Diverging Lenses

21 Thin Lenses Converging and Diverging Lenses Principle Rays

22 Thin Lenses Multiple Lens Systems How do you locate the final image? Where is the final image?

23 A- An object is placed 20 cm in front of a converging lens of focal length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image? Solution: P=20 cm, f=10 cm the same size, real image

24 An object is placed 5 cm in front of a converging lens of focal length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image? Solution: P=5cm, f=10cm q=-10 cm Virtual image, as viewed from the right, the light appears to be coming from the (virtual) image, and not the object. Magnification = +2

25 An object is placed 8 cm in front of a diverging lens of focal length 4 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image? Solution: P=8 cm, f=-4cm(concave) The image is upright, virtual, smaller


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